explicit nikulin configurations on kummersarti/newnikulinconfigurations.pdfexplicit nikulin...

EXPLICIT NIKULIN CONFIGURATIONS ON KUMMERSURFACES

XAVIER ROULLEAU, ALESSANDRA SARTI

Abstract. A Nikulin configuration is the data of 16 disjoint smoothrational curves on a K3 surface. According to results of Nikulin, theexistence of a Nikulin configuration means that the K3 surface is aKummer surface, moreover the abelian surface from the Kummer struc-ture is determined by the 16 curves. A classical question of Shioda isabout the existence of non isomorphic Kummer structures on the sameKummer K3 surface. The question was studied by several authors, andit was shown that the number of non-isomorphic Kummer structuresis finite, but no explicit geometric construction of such structures wasgiven. In the paper [15], we constructed explicitly non isomorphic Kum-mer structures on some Kummer surfaces. In this paper we generalizethe construction to Kummer surfaces with a weaker restriction on thedegree of the polarization and we describe some cases where the previousconstruction does not work.

1. Introduction

A (projective, as always in this paper) Kummer surface is obtained asthe desingularization of the quotient of an abelian surface by an involutionwith 16 isolated fixed points. It is well known that Kummer surfaces are K3surfaces and that their Picard number is at least 17, the rank 17 sub-groupbeing generated by the 16 rational curves in the resolution of the 16 nodesand by the polarization. In [12], Nikulin showed the converse, i.e. that a K3surface containing 16 disjoint smooth rational curves (or (−2)-curves) is theKummer surface associated to an abelian surface. Let X be a K3 surface; wecall a Kummer structure onX an abelian surface A (up to isomorphism) suchthat X ' Km(A), and we call a Nikulin configuration a set of 16 disjointsmooth rational curves on X. By the result of Nikulin we have a bijection:

{Kummer structures} ←→ {Nikulin configurations}/autoIn 1977, see [21, Question 5], T. Shioda raised the following question :Is it possible to have non-isomorphic abelian surfaces A and B, such that

Km(A) and Km(B) are isomorphic?Shioda and Mitani in [10, Theorem 5.1] answer negatively the question if

ρ(Km(A)) = 20, where ρ(Km(A)) is the Picard number of Km(A), i.e. the

2000 Mathematics Subject Classification. Primary: 14J28 ; Secondary: 14J50, 14J29,14J10.

Key words and phrases. Kummer surfaces, Nikulin configurations.1

2 XAVIER ROULLEAU, ALESSANDRA SARTI

rank of the Néron-Severi group of Km(A). The answer is also negative ifA is a generic principally polarized abelian surface, i.e. A is the jacobian ofa curve of genus 2 and ρ(A) = 1. Then in [7, Theorem 1.5], Gritsenko andHulek answered positively the question. They showed that if A is a generic(1, t)-polarized abelian surface with t > 1 then the abelian surface A and itsdual A, though not isomorphic, satisfy Km(A) ∼= Km(A). In [9, Theorem0.1], Hosono, Lian, Oguiso and Yau by using lattice theory showed that thenumber of Kummer structures is finite and for each integer N ∈ N∗ theyconstruct a Kummer surface of Picard number 18 with at least N Kummerstructures. Moreover in [13, Example 4.16], Orlov showed that if A is ageneric abelian surface (i.e. ρ(Km(A)) = 17) then the number of Kummerstructures is 2ν where ν is the number of prime divisors of 1

2M2, where M

is an ample generator of the Néron-Severi group of A. For example if A isprincipally polarized we have that M2 = 2 so that ν = 0 and we find againthe fact that in this case there is only one Kummer structure. Observe thatν can be also defined as the number of prime divisors of 1

4L2, where L is the

polarization induced by M on Km(A), (in particular L is orthogonal to the16 rational curves ; it is easy to see that by changing the 16 rational disjointcurves, the number ν does not change).

In [15, Theorem 1], we constructed explicit examples of two Nikulin con-figurations C, C′ such that the abelian surfaces A and A′ associated to thesetwo configurations are not isomorphic. These examples are for generic Kum-mer surfaces, such that the orthogonal complement of the 16 rational curvesin C is generated by a class L such that L2 = 2k(k + 1) for some integer k(we give a motivation for this restriction in the Appendix of this paper).

The main goal of this paper is to provide a generalization of that resultto other Kummer surfaces. For that aim, let t ∈ N be an integer and let Xbe a generic Kummer surface with a Nikulin configuration C such that theorthogonal complement of the 16 (−2)-curves A1, . . . , A16 in C is generatedby L with L2 = 4t. A class C of the form C = βL − αA1 with β ∈ N∗ hasself-intersection C2 equals to −2 if and only if the coefficients (α, β) satisfythe Pell-Fermat equation

α2 − 2tβ2 = 1.

There is a non-trivial solution if and only if 2t is not a square. Let ussuppose that this is the case. Then there exists a so-called fundamentalsolution which we denote by (α0, β0). Our main result is as follows:

Theorem 1. Suppose that β0 is even. Then β0L − α0A1 is the class ofan irreducible (−2)-curve A′1. That curve is disjoint from A2, . . . , A16 andthe Nikulin configurations C =

∑16i=1Ai and C′ = A′1 +

∑16i=2Ai define two

non-isomorphic Kummer structures on the Kummer surface X.

Observe that our result covers 34 of the cases (see Lemma 6 for a precise

meaning of that affirmation and also the table in the Appendix).One could also rise a weaker question than Shioda’s question by asking

if Km(A) ∼= Km(B) are then A and B isogenous ? The answer is positive

EXPLICIT NIKULIN CONFIGURATIONS ON KUMMER SURFACES 3

and the result was surely known, but we could not find an explicit proof inthe literature, hence we recall it in Section 2 and we show how it can beobtained as a direct consequence of a result of Stellari [22, Theorem 1.2]. Inthe rest of the paper, we point out why the construction in Theorem 1 andin [15, Theorem 1] can not work for β0 odd, moreover we study examples ofNikulin configurations in the case that β0 is odd or 2t is a square.Acknowledgements: We thank P. Stellari for pointing out his paper

[22]. We also thank K. Hulek, H. Lange, K. Oguiso, M. Ramponi, J. Rivatand T. Shioda for useful discussions.

2. Generalizations for other polarizations

2.1. The general problem.

2.1.1. Isogenies. Before to state our results about the question of Shioda [21,Question 5], we can generalize the problem to the following question:

Given two abelian surfaces A and B such that Km(A) ∼= Km(B) are thenA and B isogenous ?

The answer is positive and certainly well known, in particular to peopleworking on derived categories on abelian surfaces, for convenience we givehere a short proof:

Proposition 2. Let A and B abelian surfaces such that the associated Kum-mer surfaces are isomorphic, then A and B are isogenous abelian surfaces.

Proof. Since Km(A) ∼= Km(B) then the derived categories Db(Km(A)) andDb(Km(B)) are equivalent, then by [22, Theorem 1.2], the abelian surfacesare isogenous. �

2.1.2. Notations and known results on the Néron-Severi group of a Kummersurface. Let t ∈ N be an integer and let B be a generic Abelian surfacewith polarization M such that M2 = 2t. Let X = Km(B) be the associatedKummer surface and L be the line bundle corresponding to M , so thatL2 = 4t. Let A1, . . . , A16 be the 16 disjoint (−2)-curves on X that areresolution of the quotient B/[−1]. By [11, Proposition 3.2], [6, Proposition2.6], corresponding to the polarization M on B, there is a polarization L onKm(B) such that

L2 = 4t

and LAi = 0, i ∈ {1, . . . , 16}. The Néron-Severi group of X = Km(B)satisfies:

ZL⊕K ⊂ NS(X),

whereK denotes the Kummer lattice (the lattice generated by the 16 disjoint(−2)-curves Ai) which is a negative definite lattice of rank 16 and discrim-inant 26. For B generic among polarized Abelian surfaces rk(NS(X)) = 17and NS(X) is an overlattice of index two of ZL ⊕ K which is describedprecisely in [6, Theorem 2.7], in particular we will use the following result:


Lemma 3. ([6, Remarks 2.3 & 2.10]) An element Γ ∈ NS(X) has the formΓ = αL −

∑βiAi with α, βi ∈ 1

2Z. If α or βi for some i is in 12Z \ Z, then

at least 4 of the βj’s are in 12Z \ Z, if moreover α ∈ Z, at least 8 of the βj’s

are in 12Z \ Z.

2.1.3. The Pell-Fermat equation and construction of (−2)-classes. We arelooking to a polarization L′ and a class A′1 of the form

A′1 = βL− αA1

L′ = bL− aA1,

with α, β, a, b ∈ N \ {0} such that one has A′21 = −2, L′A′1 = 0 and L′2 =L2 = 4t. These three conditions are respectively

(2.1)α2 − 2tβ2 = 1

2tbβ = aαa2 = 2t(b2 − 1)

,

the first express that A′1 is a (−2)-class, the second that this (−2)-class isdisjoint from the polarisation L′, which must have the same square as L, thethird that L2 = L′2. We will use that last condition to show that A′1 can berepresented by an irreducible curve.

Proposition 4. There are solutions to the three equations (2.1) if and onlyif 2t is not a square.In that case, if (α, β) is a solution of the first equation in (2.1), one has

(a, b) = (2tβ, α).

Proof. Suppose that 2t = u2 with u ∈ N, then the first equation is equivalentto

(α− uβ)(α+ uβ) = 1

and then α − uβ = ±1, α + uβ = ±1, which has no integer solutions sinceα, β ∈ N \ {0}.

Let us suppose now that 2t is not a square. By the theory of the Pell-Fermat equations, there exists a solution to the first equation of (2.1). Let(α, β) be such a solution, which we can suppose with α > 0, β > 0. Byreplacing a = 2tβαb in the third equation, one gets

4t2b2β2 = 2tα2(b2 − 1),

which is equivalent tob2(α2 − 2tβ2) = α2,

since α2 − 2tβ2 = 1 and we search solutions with a > 0, b > 0, we obtainb = α. Then by the third equality, we get a2 = 2t(α2 − 1) and equalityα2 − 1 = 2tβ2 implies a = 2tβ, therefore we obtain that

(a, b) = (2tβ, α),


and finally the matrix(b β−a −α

)=

(α β−2tβ −α

)which is the transformation L → L′, A1 → A′1 in base (L,A1) has determi-nant −1 and order 2. �

2.2. Cases when 2t is not a square : the Pell-Fermat equation α2 −2tβ2 = 1. For any t ∈ N the Pell-Fermat equation

(2.2) α2 − 2tβ2 = 1

has a non-trivial solution if and only if 2t is not a square. Then there exists afundamental solution (α0, β0) ∈ N, such that for every other solution (α, β)there exists k ∈ Z with α+ β

√2t = ±(α0 +

√2tβ0)

k .

Remark 5. We observe that for a solution (α, β) of equation (2.2), the integerα is necessarily odd.

For t a positive integer such that 2t is not a square, we denote by (α0, β0)the fundamental solution of α2 − 2tβ2 = 1. The following result shows thatthe density of integers t such that β0 is even is at least 3

4 (we thank JoëlRivat for useful discussions on the following):

Lemma 6. a) Suppose that t 6= 0 mod 4. Then β0 is even.b) There is an infinite number of integers s such that the fundamental solu-tion (α0, β0) of α2 − 8s2β2 = 1 has odd β0.c) There is an infinite number of integers s such that the fundamental solu-tion (α0, β0) of α2 − 8s2β2 = 1 has even β0.

Proof. Let (α, β) be a solution of equation α2− 2tβ2 = 1. Suppose that β isodd. Then

β = ±1,±3 mod 8,

and one has β2 = 1mod 8. Since α2 − 2tβ2 = 1, one has α2 = 1 + 2t mod8. Since α is also odd, α2 = 1mod 8, thus 2t = 0 mod 8 and thereforet = 0 mod 4. That proves part a).

Let (x1, y1) be the fundamental solution of x2 − 2ty2 = 1. For n ∈ Z, theintegers ±xn,±yn defined by

xn + yn√

2t = (x1 + y1√

2t)n

are the solutions of equation x2 − 2ty2 = 1. Using that for any Pell-Fermatequation, the sequence (yn)n≥1 is strictly increasing, we see that the funda-mental solution of

x2 − 2ty2ny2 = 1

is (xn, 1). Using part a), we remark that always ty2n = 0 mod 4. Take nowt = 4, we therefore obtain result b). For n even, yn is even; let zn be suchthat yn = 2zn. The fundamental solution of

x2 − 2tz2ny2 = 1

is (xn, 2) ; taking t = 4 as in the previous case, one obtains result c). �


Example 7. For 1 ≤ s ≤ 100 such that 8s is not a square (i.e. s 6∈{2, 8, 18, 32, 50, 72, 98}), the fundamental solution (α0, β0) of equation α2 −8sβ2 = 1 is such that β0 is even if and only if s is in

{7, 9, 14, 23, 30, 31, 33, 34, 46, 47, 56, 57, 62, 63, 69, 71, 73, 75, 77, 79, 81, 82, 89, 90, 94}.

2.3. The β0 odd case. Let (α0, β0) be the fundamental solution of equation(2.2). Let us suppose that β0 is odd and let us define

A′1 = β0L− α0A1,

which is a (−2)-class. Then

Proposition 8. The (−2)-class A′1 = β0L − α0A1 cannot be the class of airreducible rational curve.

Proof. Suppose that A′1 is irreducible. Then we have two Nikulin configura-tions

C =16∑i=1

Ai, C′ = A′1 +16∑i=2

Ai.

Since both configurations are 2-divisible, the divisor A1 + A′1 is 2-divisibleand

1

2(A1 +A′1) =

β02L− α0 − 1

2A1

is an integral class. Since α0 is odd and β0 is odd too we get that L2 ∈ NS(X)

which is not possible by Lemma 3 and since we assume that L is primitivein NS(X). �

We will come back to this case in Subsection 3.1 with an example whenβ0 is odd.

2.4. The β0 even case. Let (α0, β0) be the fundamental solution of equa-tion (2.2). We assume in this section that β0 is even and we define as inSection 2.2 the classes:

A′1 = β0L− α0A1, L′ = α0L− 2tβ0A1.

One has A′21 = −2, L′A′1 = 0, L′2 = L2 = 4t.

Proposition 9. Suppose that β0 is even. The class L′ is big and nef andthe classes A′1, A2 . . . , A16 are the only (−2)-classes contracted by L′.

Proof. Let

Γ = uL−16∑i=1

viAi

be a (−2)-curve. One has∑v2i − 2tu2 = 1. Suppose that

ΓL′ ≤ 0,

this is equivalent touα0 ≤ v1β0,


in other words u ≤ β0α0v1, thus

∑v2i = 2tu2 + 1 ≤ 2t

β20

α20v21 + 1 and therefore

using the relation α20 − 2tβ20 = 1, one obtains∑

i≥2v2i ≤ 1− v21

α20

.

Apart from the trivial cases of curves Γ = Ai, one can suppose u > 0. Ifv1 >

12α0 then

∑i≥2 v

2i <

34 , this is impossible unless v1 = α0, u = β0 and

then Γ = A′1. Thus by Lemma 3 one can suppose that

0 < v1 ≤1

2α0

(if v1 = 0 then u = 0, which we excluded) and up to permutation of theindices v2 = v3 = v4 = 1

2 (since∑

i≥2 v2i < 1 and by the structure of the

Néron-Severi group as described in Lemma 3). The relation∑v2i −2tu2 = 1

is now v21 − 2tu2 = 14 , which is

(2v1)2 − 2t(2u)2 = 1.

Defining V = 2v1 and U = 2u, we see that (U, V ) are integers and aresolutions of the Pell-Fermat equation β2 − 2tα2 = 1. Since by hypothesis(α0, β0) is the primitive solution, one has α0 ≤ U = 2v. Since on the otherhand we know that v1 ≤ 1

2α0, we see that v1 = 12α0 and u = 1

2β0. Therefore

ΓL′ ≤ 0

if and only if ΓL′ = 0 and Γ = 12(β0L−α0A1 −A2 −A3 −A4). Moreover in

order for Γ to be in NS(X) the integer β0 must be odd, which is impossibleby our assumption. In any cases that divisor L′ is big and nef. We thusobtained that if β0 is even, then the only (−2)-classes Γ such that ΓL′ = 0are A′1, A2, . . . , A16. �

Let us prove the following result:

Proposition 10. Suppose that β0 is even. The line bundle 3L′ (where L′ =α0L − 2tβ0A1) defines a morphism φ3L′ : X → PN which is birational ontoits image and contracts exactly the divisor A′1 = β0L − α0A1 and the 15(−2)-curves Ai, i ≥ 2.

Proof. By [14, Section 3.8] either |3L′| has no fixed part or 3L′ = aE + Γ,where |E| is a free pencil, and Γ a (−2)-curve with EΓ = 1. However ifEΓ = 1, then 3L′E = aE2 + 1, but since E2 = 0 this is impossible. Thus|3L′| has no fixed part; moreover by [19, Corollary 3.2], it has then no basepoints.

Let us prove that the morphism φ3L′ has degree one, i.e. that |3L′| isnot hyperelliptic (see [19, Section 4]). By loc. cit., |3L′| is hyperelliptic ifthere exists a genus 2 curve C such that 3L′ = 2C or there exists an ellipticcurve E such that (3L′)E = 2. Suppose we are in the first case. SinceC2 = 2, one has 9 · 4t = 8, which is impossible. The second alternative is


also readily impossible. Thus the morphism φ3L′ has degree one. Moreoversince 3L′A′1 = 3L′A2 = · · · = 3L′A16 = 0, the 16 divisors are contracted byφ3L′ . �

We obtain:

Corollary 11. Suppose that β0 is even. The divisor A′1 is an irreducible(−2)-curve.

Proof. Since A′21 = −2 and LA′1 ≥ 0, by Riemann-Roch Theorem we canassume it is effective. Let B be one of the divisors A′1, A2, . . . , A16. One has3L′B = 0, thus the linear system |3L′| contracts B to a singular point. Sincethe Picard number of the K3 surface X = Km(B) is 17, that singularitymust be a node and therefore A′1 is irreducible. �

2.5. Two Kummer structures in case β0 even. Let (α0, β0) be the fun-damental solution of equation (2.2). We suppose that β0 is even. Let usprove that the Nikulin configurations

C =16∑i=1

Ai, C′ = A′1 +16∑i=2

Ai

defines two distinct Kummer structures, by [15, Proposition 21] this is equiv-alent to prove the following result:

Theorem 12. Suppose that t ≥ 2. There is no automorphism f of X sendingthe configuration C =

∑16i=1Ai to the configuration C′ = A′1 +

∑16i=2Ai.

In order to prove Theorem 12 let us suppose that such an automorphismf exists. The group of translations by the 2-torsion points on B acts on X =Km(B) and that action is transitive on the set of curves A1, . . . , A16. Thusup to changing f by f ◦t (where t is such a translation), one can suppose thatthe image of A1 is A′1. Then the automorphism f induces a permutation ofthe curves A2, . . . , A16. The (−2)-curve A′′1 = f2(A1) = f(A′1) is orthogonalto the 15 curves Ai, i > 1 and therefore its class is in the group generatedby L and A1. We can write the (−2)-class in NS(X) as A′′1 = λA1 + µL forcoefficients λ, µ ∈ Z . The integers λ, µ satisfy the Pell-Fermat equation

(2.3) λ2 − 2tµ2 = 1.

Let us prove:

Lemma 13. Let C = λA1 +µL be an effective (−2)-class. Then there existsu, v ∈ N such that λA1+µL = uA1+vA′1, in particular the only (−2)-curvesin the lattice generated by L and A1 are A1 and A′1.

Proof. If (λ, µ) is a solution of equation (2.3), then so are (±λ,±µ). We saythat a solution is positive if λ ≥ 0 and µ ≥ 0. Let us identify Z2 with Z[

√2t]

by sending (λ, µ) to λ+µ√

2t. The solutions of equation 2.3 are units of thering Z[

√2t]. Let α0 + β0

√2t (α0, β0 ∈ N∗) be the fundamental solution to


equation (2.3). The solutions with positive coefficients are the elements ofthe form

λm + µm√

2t = (α+ β√

2t)m, m ∈ N.An effective (−2)-class C = λA1 + µL either equals A1 or satisfies CL > 0and CA1 > 0, therefore b > 0 and a < 0. Thus if C 6= A1, there exists msuch that C = −λmA1 + µmL. Since A′1 = β0L − α0A1 corresponds to thefundamental solution of equation (2.3), we have L = 1

β0(A′1 + α0A1) and we

obtain

C = −λmA1 +µmβ0

(A′1 + α0A1) =µmβ0A′1 + (

α0

β0bm − λm)A1

and the Lemma is proved if the coefficients um = µmβ0

and vm = α0β0µm − λm

are both positive and in Z. Using the fact that

λm+1 + µm+1

√2t = (α0 +

√2tβ0)(λm + µm

√2t),

we obtainλm+1 = α0λm + 2tβ0µmµm+1 = α0µm + β0λm

.

Then we compute that

um+1 =µm+1

β0= α0

µmβ0

+ λm, vm+1 =α0

β0µm+1 − λm+1 =

µmβ0

and by induction we conclude that um, vm are in N for any m ≥ 1. �

Therefore we see that A′′1 = A1 i.e. f permutes A1 and A′1. Let uscomplete the proof of Theorem 12:

Proof. The class f∗L is orthogonal to the rank 16 lattice generated byA′1, A2, . . . , A16, thus this is a multiple of the class L′ = β0L − α0A1 whichhas the same property. Since both classes have the same self-intersectionand are effective, we get f∗L = L′; by the same reasoning, since f∗A′1 = A1,we get f∗L′ = L. By [6, Proposition 4.3], the divisor

D = 2L− 1

2

∑i≥1

Ai

is ample, thus f∗D = 2L′− 12(A′1+

∑i≥2Ai) is also ample and so is D+f∗D.

Moreover D + f∗D is invariant by f , thus by [8, Proposition 5.3.3], theautomorphism f has finite order. Up to taking a power of it, one can supposethat f has order 2m for somem ∈ N∗. Supposem = 1, i.e. f is an involution.Then the integral class

1

2(A1 +A′1) =

β02L− α0 − 1

2A1

(recall that β0 is even and α0 is odd) is fixed; there are curves Ai, i > 1 suchthat f(Ai) = Ai (say s of such curves; necessarily s is odd) and f permutesthe remaining curves Aj by pairs (there are s′ = 1

2(15− s) such pairs). LetΓ be the lattice generated by the classes Ai fixed by f , by Aj + f(Aj) if


f(Aj) 6= Aj and by 12(A1 +A′1). It is a finite index sub-lattice L of NS(X)f ,

the fix sub-lattice of the Néron-Severi group. The discriminant group of L is

Z/(α0 − 1)Z× (Z/2Z)s × (Z/4Z)s′.

Since in NS(X) there is at most a coefficient 12 on L, the discriminant

of NS(X)fcontains a subgroup isomorphic Z/[12(α0 − 1)Z]. If f was non-symplectic, then M = NS(X)f would be a 2-elementary lattice (see [1]; itmeans that the discriminant group M∗/M ' (Z/2Z)h for some positive in-teger h). But if 1

2(α0 − 1) > 2 this is impossible, and therefore f must besymplectic in that case. One has 1

2(α0 − 1) ≤ 2 if and only if α0 ≤ 5. Thetriplets (α0, β0, t) such that (α0, β0) is the fundamental solutions of equationα2 − 2tβ2 = 1 with β0 even and α0 ≤ 5 are

(3, 2, 1), (5, 2, 3).

The case (α0, β0, t) = (3, 2, 1) is excluded since we suppose t ≥ 2. The case(α0, β0, t) = (5, 2, 3) has been studied and excluded in [15, Theorem 19].

We therefore proved that for any t > 1, f must be symplectic.A symplectic automorphism acts trivially on the transcendental lattice TX ,which in our situation has rank 5. Therefore the trace of f on H2(X,Z)equals 6 + s > 6. But the trace of a symplectic involution equals 6 (see e.g.[20, Section 1.2]). This is a contradiction, thus f cannot have order 2 andthe integer m (such that the order of f is 2m) is larger than 1.

The automorphism g = f2m−1 has order 2 and g(A1) = A1, g(A′1) = A′1,

thus g(L) = L. There are curves Ai, i > 1 such that f(Ai) = Ai (say sof such curves, s is odd since A1 is fixed) and the remaining curves Aj arepermuted 2 by 2 (there are s′ = 1

2(15−s) such pairs). Let similarly as aboveL′ be the sub-lattice generated by L,A1 and the fix classes Ai, Aj + g(Aj).It is a finite index sub-lattice of NS(X)g and its discriminant group is

Z/4tZ× (Z/2Z)s+1 × (Z/4Z)s′.

By the same reasoning as before, the automorphism g must be symplecticas soon as t > 1. However the trace of g is 8 + s > 6, thus g cannot besymplectic either. Therefore we conclude that such an automorphism f doesnot exist. �

3. further examples

3.1. An example of a Nikulin configuration when β0 is odd. Let usstudy the t = 4 case. Then the fundamental solution (α0, β0) equals (3, 1).This is the first case with β0 odd (see Table in the Appendix). We have

A′1 = L− 3A1, L′ = 3L− 8A1


and we already know that A′1 is not irreducible. In order to understandbetter what is happening, let us define

A′′1 = 12(L− 3A1 −A2 −A3 −A4)

A′′2 = 12(L−A1 − 3A2 −A3 −A4)

A′′3 = 12(L−A1 −A2 − 3A3 −A4)

A′′4 = 12(L−A1 −A2 −A3 − 3A4)

,

where the classes A2, A3, A4 are chosen so that the classes A′′j exists in NS(X)

(which is possible by [6, Theorem 2.7], since t = 0 mod 2). We compute thatthese are (−2)-classes i.e. A′′2j = −2. Moreover we have

A′1 = 2A′′1 +A2 +A3 +A4 and A′′1L′ = 0,

(but A′′iL′ 6= 0, for i = 2, 3, 4). Let us also define

L1 = 3L− 4(A1 +A2 +A3 +A4).

We remark that L21 = L2 = 16, L1A

′′i = 0 and A′′iA

′′j = 0 for i 6= j in

{1, 2, 3, 4}.

Lemma 14. The class L1 is big and nef. Moreover if Γ is an effective (−2)-class, we have L1Γ ≥ 0 and L1Γ = 0 if and only if Γ is one of the classesA′′1, ..., A

′′4, A5, . . . , A16.

Proof. Let

Γ = aL−16∑i=1

biAi

be an effective (−2)-class (thus∑b2i − 8a2 = 1). One has

L1Γ ≤ 0

if and only if6a ≤ b1 + b2 + b3 + b4.

Suppose Γ 6∈ {A5, . . . , A16}. Then a > 0, bi ≥ 0 and equation L1Γ ≤ 0 isequivalent to

a2 ≤ 1

36(b1 + · · ·+ b4)

2.

Using(b1 + · · ·+ b4)

2 ≤ 4(b21 + · · ·+ b24) ≤ 4(b21 + · · ·+ b216)

we get

a2 ≤ 1

9(b21 + · · ·+ b216)

and since∑b2i = 8a2 + 1, we have

a2 ≤ 1

9(8a2 + 1),

thus a2 ≤ 1 and a ∈ {12 , 1}. Suppose that a = 12 . Then

∑i=16i=1 b2i = 3 and

either there are 12 bi’s equal to 12 or (up to permutation of the indices) b1 = 3

2 ,b2 = b3 = b4 = 1

2 . The first case is impossible since 3 = 6a > b1+b2+b3+b4.


The second case corresponds to A′′1, . . . , A′′4, and then L1A′′j = 0.

It remains to study the case a = 1, then∑b2i = 9 and

6 ≤ b1 + b2 + b3 + b4.

That implies bi ≤ 52 . Up to permutation we can suppose that the largest

bi with i ∈ {1, 2, 3, 4} is b1. Suppose b1 = 52 , then

∑i≥2 b

2i = 11

4 andb2 + b3 + b4 ≥ 7

2 . One can suppose that b2 is the largest among b2, b3, b4,then there are two cases : b2 = 2 or b2 = 3

2 . The first case is impossiblesince one would obtain

∑i≥2 b

2i >

114 . Suppose b2 = 3

2 , then b23 + b24 = 1

2 andb3 + b4 ≥ 2, but this is also impossible.Suppose that b1 = 2. Then

∑i≥2 b

2i = 5 and b2 + b3 + b4 ≥ 4. The largest

bi among b2, b3, b4 (say it is b2) is 2 or 32 . If b2 = 2, then b23 + b24 = 1

and b3 + b4 ≥ 2, which is impossible. If b2 = 32 , then b3 + b4 ≥ 5

2 andb23 + b24 = 11

4 ,thus b3 = 32 and b4 ≥ 1 gets a contradiction.

It remains b1 = 32 , but then b2 = b3 = b4 = 3

2 . That implies bj = 0 for j ≥ 5.But L− 3

2(A1 +A2 +A3 +A4) is not in the Néron-Severi group of the surface(see Lemma 3).We thus proved that the only effective (−2)-classes Γ such that L1Γ ≤ 0 areA′′1, . . . , A

′′4, A5, . . . , A16 and moreover L1Γ = 0 for these classes. Thus L1 is

nef and big. �

As before, one can prove that the linear system 3L1 define a degree 1morphism which contracts A′′1, . . . , A′′4, A5, . . . , A16 onto singularities. Sincewe assume that the Kummer surface is generic, it has Picard number 17 andwe conclude that the divisors A′′1, . . . , A′′4 are irreducible. Therefore:

Corollary 15. The 16 (−2)-curves A′′1, . . . , A′′4, A5, . . . , A16 form a Nikulin

configuration C′ on the K3 surface X. The (−2)-class A′1 is not irreducibleand A′1 = 2A′′1 +A2 +A3 +A4.

Remark 16. i) One can check that the class L′ is big and nef; the image ofX by the linear system |3L′| is a surface with 12 nodal singularities and oneD4 singularity obtained by contracting A′′1, A2, A3, A4.ii) We do not know yet if C′ is another Kummer structure on the Kummersurface X, we intend to study that problem in a forthcoming paper.

3.2. An example of a Nikulin configuration when 2t is a square. Letus consider the case t = 2 i.e. A is a (1, 2)-polarized abelian surface. Then 2tis a square and the method in Section 2.4 do not apply. We start by recallingthe following

Remark 17. Since t is even, by [6, Theorem 2.7 and Remark 2.10], we canlabel the 16 curves (−2)-curves Aj so that the classes

12(L−A1 −A2 −A3 −A4),

12(L−A5 −A6 −A7 −A8),

12(L−A9 −A10 −A11 −A12),

12(L−A13 −A14 −A15 −A16)

are contained in NS(X).


We take that labelling and we define the classes

F1 =1

2(L−A1−A2−A3−A4), F2 =

1

2(L−A5−A6−A7−A8) ∈ NS(X).

For j ∈ {1, ..., 4}, we define

Bj = F2 −Ajand for j ∈ {5, ..., 8}, we define

Bj = F1 −Aj .These are (−2)-classes; they are effective since LBj > 0. We check moreoverthat

BjBk = −2δjk,

where δjk is the Kronecker symbol. Let us prove the following result:

Proposition 18. The classes B1, . . . , B8, A9, . . . , A16 are 16 disjoint (−2)-curves.

Proof. We have BkAj = 0 for k ∈ {1, . . . , 8} and j ∈ {9, . . . , 16}. It remainsto prove that B1, . . . , B8 are irreducible. We compute that F 2

1 = 0 = F 22 ,

F1F2 = 2. The linear system |Fk| defines a fibration ψk : X → P1. SinceF1Aj = 1 for j ∈ {1, ..., 4}, the fibration ψ1 has connected fibers. By thesame kind of argument so is ψ2. For k ∈ {5, ..., 16}, let us define

Ck = F1 −Ak(so that in fact Bk = Ck for k ∈ {5, ..., 8}). The divisor Ck is an effective(−2)-class and the 12 divisors

Ck +Ak, k ∈ {5, . . . , 16}are distinct singular fibers of ψ1, with AkCk = 2. We now use [3, Proposition11.4, Chapter III]: the Euler characteristic of X (equal to 24) is the sum∑

s e(fs) of the Euler numbers of all the singular fibers. By the Kodairaclassification of singular fibers of elliptic fibrations (see e.g. [3, Table 3,Chapter V, Section 7]), the singular fibers fs = Ck + Ak for k ≥ 5 satisfye(fs) ≥ 2. Moreover, by the above cited Table, a singular fiber fs containinga smooth rational curve satisfies e(fs) = 2 if and only if it is the union of two(−2)-curves D1, D2 with D1D2 = 2 and meeting transversally. Computingthe Euler characteristic of X, we see that necessarily e(Ck+Ak) = 2, for k ∈{5, . . . , 16} and therefore the curves Bk = Ck k ∈ {5, ..., 8} are irreducible(−2)-curves. We proceed in a similar way with ψ2 for the curves Bk withk ∈ {1, ..., 4}, thus we obtain the result. �

Remark 19. i) By the Proposition 18, we see that the elliptic fibration definedby F1 contains 12 fibers of type I2. By general results on elliptic K3 surfaces,the rank ρ of the Néron-Severi group is 14 = 12 + 2 plus the rank of theMordell-Weil group, which is the group generated by the zero section (wecan take A1 as zero section) and the sections of infinite order. Since we knowthat ρ = 17 we get that the rank of the Mordell-Weil group is three. That


group contains the sections A2, A3 and A4. The remark is similar for thefibration defined by F2.ii) On the K3 surface X we have two Nikulin configurations

C =16∑i=1

Ai, C′ =8∑i=1

Bi +16∑i=9

Ai.

We do not know if these configurations define two Kummer structures on X.We intend to come back on the subject later.iii) It is also possible to check that the divisor L′ = 3L− 2(A1 + · · ·+A8) isbig and nef, L′2 = 8 and L′Γ = 0 for an effective (−2)-class Γ if and only ifΓ is in {B1, . . . , B8, A9, . . . , A16}.

Appendix

Why it was natural to study the case t = 12k(k+1) in the paper [15].

Since α2 = 1+2tβ2, the integer α is odd. Let k ∈ N be such that α = 2k+1(then one has A1A

′1 = 4k+2). The integer β is then solution of the equation

(2k + 1)2 − 2tβ2 = 1,

which is equivalent to

tβ2 = 2k(k + 1).

Then

a = 2tβ, b = 2k + 1

are solutions of the three conditions in (2.1). Since a2 = 2t(b2− 1), one gets

(3.1) a2 = 2t · 4k(k + 1).

Thus 2t ·4k(k+1) must be the square of an integer and it is therefore naturalto define

t =1

2k(k + 1).

Then one computes easily that a = 2k(k+ 1) and β = 2. That was the caseswe studied in [15].

A table. We resume in the following table the solutions of the Pell’s equa-tion α2 − 2tβ2 = 1 for t ≤ 30. Recall that there are non-trivial solutions ifand only if 2t is not a square. Observe that when 2t = k(k+ 1) the minimalsolution is (2k + 1, 2), these correspond to Nikulin configurations studiedin the paper [15], we put a ∗ close to these cases. Moreover we put a boxaround the cases with β odd, which are left out in this paper.


Table 1. First solutions of Pell’s equation

2t 2* 4 6* 8 10 12* 14 16 18 20* 22 24 26 28 30*α 3 - 5 3 19 7 15 - 17 9 197 5 51 127 11β 2 - 2 1 6 2 4 - 4 2 42 1 10 24 22t 32 34 36 38 40 42* 44 46 48 50 52 54 56* 58 60α 17 35 - 37 19 13 199 24335 7 99 649 485 15 19603 31β 3 6 - 6 3 2 30 3588 1 14 90 66 2 2574 40

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Xavier Roulleau,Aix-Marseille Université, CNRS, Centrale Marseille,I2M UMR 7373,13453 Marseille, [email protected]

Alessandra SartiUniversité de PoitiersLaboratoire de Mathématiques et Applications,UMR 7348 du CNRS,TSA 6112511 bd Marie et Pierre Curie,86073 POITIERS Cedex 9,France

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